Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Maximum number that can be display on Seven Segment Display using N segments in C++
Given the task is to find the maximum number that can be displayed using N segment on ant number of seven segment display.
Let’s now understand what we have to do using an example −
Input − N=5
Output − 71
Explanation − The largest number will be displayed as follows on the seven segment display −
Input − N=6
Output − 111
Approach used in the below program as follows
The following situation can be divided into 3 case −
Case 1 −
If N is 0 or 1, then it is not possible to display any number.
Case 2 −
If N is odd. Then the numbers that can be displayed with odd number of segments are 2, 3, 5, 7 and 8 out of which 7 takes the least number of segments, that is, 3. Therefore 7 is the most significant digit in this case.
Case 3 −
If N is even. Then the numbers that can be displayed with odd number of segments are 0, 1, 4, 6, 9 and 8 out of which 1 takes the least number of segments, that is, 2. Therefore 1 is the most significant digit in this case.
In function MaxNumber() first check for the base condition using
if (N == 1 || N == 0)
Then using another if statement, check if N is even. If so then print “1” as it is the most significant digit in the even case and call MaxNumber(N - 2) as 2 segments are being used here.
Use another if statement to check if N is odd. If so then print “7” as it is the most significant digit in the odd case and call MaxNumber(N - 3) as 3 segments are being used here.
Example
#include <iostream>
using namespace std;
void MaxNumber(int N){
//Condition to check base case
if (N == 1 || N == 0){
return;
}
//If the number is even
if (N % 2 == 0){
cout << "1";
MaxNumber(N - 2);
}
//If the number is odd
else if (N % 2 == 1){
cout << "7";
MaxNumber(N - 3);
}
}
//Main function
int main(){
int N;
N = 5;
MaxNumber(N);
return 0;
}Output
If we run the above code we will get the following output −
71
247 Views
To Continue Learning Please Login